A committee

A committee of 6 is chosen from 8 men and seven women. Find how many committees are possible if a particular man must be included.

n = 4998

C_{1} (8) = (1 8) = \frac{8 !}{1 ! ( 8 - 1 ) !} = \frac{8}{1} = 8 C_{5} (7) = (5 7) = \frac{7 !}{5 ! ( 7 - 5 ) !} = \frac{7 \cdot 6}{2 \cdot 1} = 21 C_{3} (8) = (3 8) = \frac{8 !}{3 ! ( 8 - 3 ) !} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56 C_{5} (8) = (5 8) = \frac{8 !}{5 ! ( 8 - 5 ) !} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56 a = (1 8) \cdot (5 7) = 8 \cdot 21 = 168 C_{2} (8) = (2 8) = \frac{8 !}{2 ! ( 8 - 2 ) !} = \frac{8 \cdot 7}{2 \cdot 1} = 28 C_{4} (7) = (4 7) = \frac{7 !}{4 ! ( 7 - 4 ) !} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 C_{4} (8) = (4 8) = \frac{8 !}{4 ! ( 8 - 4 ) !} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70 b = (2 8) \cdot (4 7) = 28 \cdot 35 = 980 C_{3} (7) = (3 7) = \frac{7 !}{3 ! ( 7 - 3 ) !} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 c = (3 8) \cdot (3 7) = 56 \cdot 35 = 1960 C_{2} (7) = (2 7) = \frac{7 !}{2 ! ( 7 - 2 ) !} = \frac{7 \cdot 6}{2 \cdot 1} = 21 d = (4 8) \cdot (2 7) = 70 \cdot 21 = 1470 C_{1} (7) = (1 7) = \frac{7 !}{1 ! ( 7 - 1 ) !} = \frac{7}{1} = 7 e = (5 8) \cdot (1 7) = 56 \cdot 7 = 392 C_{6} (8) = (6 8) = \frac{8 !}{6 ! ( 8 - 6 ) !} = \frac{8 \cdot 7}{2 \cdot 1} = 28 C_{0} (7) = (0 7) = \frac{7 !}{0 ! ( 7 - 0 ) !} = \frac{1}{1} = 1 f = (6 8) \cdot (0 7) = 28 \cdot 1 = 28 n = a + b + c + d + e + f = 168 + 980 + 1960 + 1470 + 392 + 28 = 4998

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Showing 1 comment:

Math student

The answer is actually 14C5=2002

1 year ago Like

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